HP 50G

Page 18-34
The confidence interval for the population variance σ
2
is therefore,
[(n-1)⋅S
2
/ χ
2
n-1,α/2
; (n-1)⋅S
2
/ χ
2
n-1,1-α/2
].
where χ
2
n-1,α/2
, and χ
2
n-1,1-α/2
are the values that a χ
2
variable, with ν = n-1
degrees of freedom, exceeds with probabilities α/2 and 1- α /2, respectively.
The one-sided upper confidence limit for σ
2
is defined as (n-1)⋅S
2
/ χ
2
n-1,1-α
.
Example 1
– Determine the 95% confidence interval for the population variance
σ
2
based on the results from a sample of size n = 25 that indicates that the
sample variance is s
2
= 12.5.
In Chapter 17 we use the numerical solver to solve the equation α = UTPC(γ,x).
In this program, γ represents the degrees of freedom (n-1), and α represents the
probability of exceeding a certain value of x (χ
2
), i.e., Pr[χ
2
> χ
α
2
] = α.
For the present example, α = 0.05, γ = 24 and α = 0.025. Solving the
equation presented above results in χ
2
n-1,α/2
= χ
2
24,0.025
= 39.3640770266.
On the other hand, the value χ
2
n-1,α/2
= χ
2
24,0.975
is calculated by using the
values γ = 24 and α = 0.975. The result is χ
2
n-1,1-α/2
= χ
2
24,0.975
=
12.4011502175.
The lower and upper limits of the interval will be (Use ALG mode for these
calculations):
(n-1)⋅S
2
/ χ
2
n-1,α/2
= (25-1)⋅12.5/39.3640770266 = 7.62116179676
(n-1)⋅S
2
/ χ
2
n-1,1-α/2
= (25-1)⋅12.5/12.4011502175 = 24.1913044144
Thus, the 95% confidence interval for this example is:
7.62116179676 < σ
2
< 24.1913044144.